AP Calculus AB Introducing Limits Quiz: 10 Questions with Explanations

A \(3\)B \(0\)C \(4\)D \(\text{does not exist}\)

The outputs approach height 4 from both sides, even though the point at x = 3 is a hole. The tempting wrong answer is to say the limit does not exist because the function is undefined there.

A it equals \(4\)B it equals \(6\)C it does not exist, because the left side approaches 4 and the right side approaches 6D it equals \(5\), the average of the two sides

The outputs from the left head toward 4 and from the right toward 6, so the one-sided limits differ and the two-sided limit does not exist. Averaging them is a common error.

A \(\lim_{x\to 1} f(x)=4\)B \(\lim_{x\to 1} f(x)\) does not existC \(\lim_{x\to 1^-} f(x)=3\)D \(\lim_{x\to 1^+} f(x)=5\)

Both pieces reach height 4 at x = 1, so the left and right limits agree. The wrong answers confuse the intercept or the formula with the value at x = 1.

A \(f(5)=9\) exactlyB \(f(x)\) gets close to \(9\) when \(x\) is close to \(5\)C \(f(x)\) gets close to \(5\) when \(x\) is close to \(9\)D \(f\) is continuous at \(x=5\)

A limit describes nearby outputs. It does not guarantee that f(5) equals 9 or that the function is continuous.

A \(\lim_{x\to a^-} f(x)\)B \(\lim_{x\to a^+} f(x)\)C \(\lim_{x\to -a} f(x)\)D \(f(a)\)

The minus sign on a means approaching from values less than a. The plus sign means from the right.

A the left and right limits are both \(3\)B the left limit is \(2\) and the right limit is \(5\)C \(f(0)\) is undefined but both sides approach \(1\)D the outputs approach \(0\) from both sides

A jump (different one-sided limits) means no two-sided limit exists. An undefined function value alone does not stop the limit from existing.

A \(3\)B \(2.5\)C it does not existD \(2\)

For x just below 3, such as 2.99, the floor is 2, so the left-hand limit is 2 even though the value at 3 is 3.

A \(12\)B \(4\)C \(8\)D \(6\)

The average rate is (16 – 4)/(4 – 2) = 6. Using f(4) alone gives 16, and 12 is the change in f without dividing by the width.

A \(10\)B \(0\)C \(5\)D \(\text{does not exist}\)

For x not equal to 5, f(x) = x + 5, so the limit is 10. The expression is 0/0 at x = 5, but that does not mean the limit is 0.

A it approaches \(0\)B it grows without bound, so the limit does not exist as a real numberC it approaches \(1\)D it approaches \(-\infty\)

Squaring makes every output positive and larger as x shrinks: f(0.1) = 100, f(0.01) = 10,000. So it grows without bound and does not approach -infinity.